Question Number 68390 by TawaTawa last updated on 10/Sep/19
![](https://www.tinkutara.com/question/9234.png)
Answered by som(math1967) last updated on 10/Sep/19
![tan(2tan^(−1) (√((2cos^2 (θ/2))/(2sin^2 (θ/2)))) )+tanθ =tan{2tan^(−1) (cot (θ/2))}+tan θ =tan{2tan^(−1) tan((π/2)−(θ/2))}+tan θ =tan(π−θ)+tanθ =−tanθ+tanθ=0](https://www.tinkutara.com/question/Q68395.png)
$${tan}\left(\mathrm{2tan}^{−\mathrm{1}} \sqrt{\frac{\mathrm{2}{cos}^{\mathrm{2}} \frac{\theta}{\mathrm{2}}}{\mathrm{2}{sin}^{\mathrm{2}} \frac{\theta}{\mathrm{2}}}}\:\right)+{tan}\theta \\ $$$$={tan}\left\{\mathrm{2tan}^{−\mathrm{1}} \left(\mathrm{cot}\:\frac{\theta}{\mathrm{2}}\right)\right\}+\mathrm{tan}\:\theta \\ $$$$={tan}\left\{\mathrm{2tan}^{−\mathrm{1}} {tan}\left(\frac{\pi}{\mathrm{2}}−\frac{\theta}{\mathrm{2}}\right)\right\}+\mathrm{tan}\:\theta \\ $$$$={tan}\left(\pi−\theta\right)+{tan}\theta \\ $$$$=−{tan}\theta+{tan}\theta=\mathrm{0} \\ $$
Commented by TawaTawa last updated on 10/Sep/19
![God bless you sir](https://www.tinkutara.com/question/Q68413.png)
$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir} \\ $$